If tan (A + B) = sqrt{3} and tan (A - B) = fr | vedclass

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(A) Given that $	an (A + B) = \sqrt{3}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $A + B = 60^{\circ} \dots (1)$.Also,$	an (A - B) = \frac{1}{\sqrt

Vedclass · Accepted Answer

$A = 45^{\circ}, B = 15^{\circ}$

Vedclass · Answer

(A) Given that $	an (A + B) = \sqrt{3}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $A + B = 60^{\circ} \dots (1)$.Also,$	an (A - B) = \frac{1}{\sqrt{3}}$.Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $A - B = 30^{\circ} \dots (2)$.Adding equations $(1)$ and $(2)$:$(A + B) + (A - B) = 60^{\circ} + 30^{\circ}$$2A = 90^{\circ}$$A = 45^{\circ}$.Substituting $A = 45^{\circ}$ in equation $(1)$:$45^{\circ} + B = 60^{\circ}$$B = 60^{\circ} - 45^{\circ} = 15^{\circ}$.Thus,$A = 45^{\circ}$ and $B = 15^{\circ}$.

If $\tan (A + B) = \sqrt{3}$ and $\tan (A - B) = \frac{1}{\sqrt{3}}$,where $0^{\circ} < A + B \leq 90^{\circ}$ and $A > B$,find the values of $A$ and $B$.

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If $\tan (A + B) = \sqrt{3}$ and $\tan (A - B) = \frac{1}{\sqrt{3}}$,where $0^{\circ} < A + B \leq 90^{\circ}$ and $A > B$,find the values of $A$ and $B$.

Similar Questions

Evaluate:$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$

Prove the following identity,where the angles involved are acute angles for which the expressions are defined:$(\operatorname{cosec} A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$

Evaluate the following:$\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$ (in $/12$)

Evaluate the following:$2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

Evaluate:$\operatorname{cosec} 31^{\circ}-\sec 59^{\circ}$